Select the unit vector in the direction of $\vec{v}=\left( -7, 6 \right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left( {-\dfrac{7}{\sqrt{85}}},{\dfrac{6}{\sqrt{85}}}\right) $ (Choice B) B $\left( {\dfrac{6}{\sqrt{85}}},{-\dfrac{7}{\sqrt{85}}}\right) $ (Choice C) C $\left( {-\dfrac{7}{\sqrt{13}}},{\dfrac{6}{\sqrt{13}}}\right) $ (Choice D) D $\left( {\dfrac{6}{\sqrt{13}}},{-\dfrac{7}{\sqrt{13}}}\right) $
Explanation: Getting started A unit vector has a magnitude (or length) of $1$. Dividing $\vec v$ by its magnitude will find a vector in the same direction as $\vec v$ but with a magnitude of $1$ : $\text{Unit vector in the direction of } \vec v = \dfrac{\vec v}{|| \vec v||}$ Finding the unit vector $\begin{aligned} \dfrac{\vec{v}}{||\vec{v}||} &= \dfrac{\left( -7, 6 \right) }{\sqrt{(-7)^2+6^2}} \\\\\\ &= \dfrac{1}{\sqrt{85}} \cdot \left( -7, 6 \right) \\\\\\ &= \left( {-\dfrac{7}{\sqrt{85}}}, {\dfrac{6}{\sqrt{85}}}\right) \end{aligned}$ Great, we found the unit vector! Just to be careful, let's check and make sure it has a magnitude of $1$. Verifying that the magnitude is $1$ $\begin{aligned}&\sqrt{\left( {-\dfrac{7}{\sqrt{85}}} \right)^2 + \left( {\dfrac{6}{\sqrt{85}}} \right)^2} \\\\\\ &= \sqrt{\dfrac{49}{85} + \dfrac{36}{85}} \\\\\\ &= \sqrt{\dfrac{85}{85}} \\\\\\ &=1 \end{aligned}$ The answer $\left( {-\dfrac{7}{\sqrt{85}}}, {\dfrac{6}{\sqrt{85}}}\right) $ Visualizing the answer: $-7$ $6$ $\dfrac{-7}{\sqrt{85}}$ $\dfrac{6}{\sqrt{85}}$ $\vec{v}$ ${\text{unit vector}}$ Notice how the unit vector points in the same direction as the original vector, but the unit vector has a magnitude of $1$.